Problem 64 - A Graphical Approach to College Algebra

Question

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to complete the solution. $$\begin{array}{l}x+9 y=-15 \\\3 x+2 y=5\end{array}$$

Step-by-Step Solution

Answer

The solution is \(x = 3\) and \(y = -2\).

Step 1: Write the Linear System in Matrix Form

The system of equations is given as: \[\begin{align*}x + 9y &= -15, \3x + 2y &= 5.\end{align*}\]This can be represented in matrix form as \(A \mathbf{x} = \mathbf{b}\), where:\[A = \begin{bmatrix}1 & 9 \ 3 & 2\end{bmatrix}, \, \mathbf{x} = \begin{bmatrix}x \ y\end{bmatrix}, \, \text{and } \mathbf{b} = \begin{bmatrix}-15 \ 5\end{bmatrix}.\]

Step 2: Calculate the Determinant of Matrix A

To apply Cramer's rule, we first need to compute the determinant of matrix \(A\). The determinant \(D\) is given by:\[D = \det(A) = \begin{vmatrix}1 & 9 \ 3 & 2\end{vmatrix} = (1)(2) - (9)(3) = 2 - 27 = -25.\]

Step 3: Compute Determinants for Cramer's Rule

Since \(D eq 0\), Cramer's Rule can be applied. To find \(x\) and \(y\), compute:\[D_x = \begin{vmatrix}-15 & 9 \ 5 & 2\end{vmatrix} = (-15)(2) - (9)(5) = -30 - 45 = -75,\]\[D_y = \begin{vmatrix}1 & -15 \ 3 & 5\end{vmatrix} = (1)(5) - (-15)(3) = 5 + 45 = 50.\]

Step 4: Solve for x and y Using Determinants

Using Cramer's Rule:\[x = \frac{D_x}{D} = \frac{-75}{-25} = 3,\]\[y = \frac{D_y}{D} = \frac{50}{-25} = -2.\]

Step 5: Verify the Solution (Optional)

To ensure correctness, substitute \(x = 3\) and \(y = -2\) into the original equations:1. \(x + 9y = 3 + 9(-2) = 3 - 18 = -15\).2. \(3x + 2y = 3(3) + 2(-2) = 9 - 4 = 5\).The solution satisfies both equations.

Key Concepts

Systems of Equations

A system of equations is a set of two or more equations with the same variables. In this exercise, the system comprises two equations:

• \(x + 9y = -15\)
• \(3x + 2y = 5\)

The objective is to find the values of the variables \(x\) and \(y\) that satisfy both equations simultaneously. Solving these systems can be done using various methods such as substitution, elimination, or matrix operations, like Cramer's Rule.
Cramer's Rule is particularly useful here because the system has the same number of equations as unknowns, making the system square. Square systems of equations can often be solved more straightforwardly using matrix operations if the determinant of the coefficient matrix is non-zero. When working with systems of equations, the result must be checked against both original equations to confirm accuracy.

Matrix Determinant

The determinant of a matrix is a special number that can provide insights into the properties of a matrix. For a 2x2 matrix \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\]the determinant is calculated as:

• \(\det(A) = ad - bc\)

In our exercise, we have the matrix:\[A = \begin{bmatrix} 1 & 9 \ 3 & 2 \end{bmatrix}\]Thus, the determinant \(D\) is \(\det(A) = (1)(2) - (9)(3) = -25\).
A non-zero determinant indicates that Cramer's Rule can be applied because it guarantees a unique solution for the system. If the determinant were zero, it would imply that the system could be either inconsistent or have infinitely many solutions, necessitating an alternative method for solving the equations.

Linear Algebra

Linear algebra provides the tools to express systems of equations compactly using matrices, allowing for more efficient computations. In this context, the system of equations is represented in the matrix equation form \(A\mathbf{x} = \mathbf{b}\), where:

• \(A\) is the matrix of coefficients.
• \(\mathbf{x}\) is a vector of variables,\[\mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}.\]
• \(\mathbf{b}\) is a vector of solutions,\[\mathbf{b} = \begin{bmatrix} -15 \ 5 \end{bmatrix}.\]

By using concepts such as matrix determinants and vectors, linear algebra provides a structured way of solving systems of equations like the one in this exercise. It enables considering the relationships between the equations as operations on vectors in a linear space. This underlying theory allows techniques like Cramer's Rule to solve equations efficiently when the system is consistent and the coefficient matrix is non-singular (non-zero determinant). Understanding these foundations can significantly simplify solving complex systems encountered in algebra.